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ASK THE PROFESSOR FORUM

Course Documents

Chapter 1 - Intro

Chapter 2 - Methods for Describing Sets of Data

Chapter 3 - Probability

Chapter 4 - Discrete Random Variables

Chapter 5 - Normal Random Variables

Chapter 6 - Sampling Distributions

Chapter 7 - Confidence Intervals

Chapter 8 - Tests of Hypothesis: One Sample

Chapter 9 - Confidence Intervals and Hypothesis Tests: Two Samples

Sample Exam I: Chapters 1 & 2

Sample Exam II: Chapters 3 & 4

Sample Exam III: Chapters 5 & 6

Sample Exam IV: Chapters 7 & 8

Ask the Professor Forum

Dear Professor, I came across two more questions as I was studying.

For Now You Try It Section 4.2 Discrete Random Variables

For Number one the part that asks if it would be unusual for someone to roll 3 or 4 sevens out of four rolls how do you know to use within 2 standard deviations of the mean? And on the test will you speficy by saying capture 75%? Also how do we know whether we use a z-score or chevy's theorem or empirical to figure out if it is unusual?

Problem 1:
A pair of six-sided die is rolled four times. We record the number of sevens rolled
over the four rolls. Calculate the standard deviation for the number of sevens
rolled. Would it be unusual for someone to roll 3 or 4 sevens out of four rolls?

Solution:
The standard deviation is: σ = 0.746 . Based on the following interval it
would be unusual to roll more than two sevens out of four rolls:
( 2 , 2 ) (0.667 2*0.746,0.667 2*0.746) ( 0.825, 2 µ σ µ σ − + = − + = − .159) .

For Problem 2:
I am not sure what information to use to go about finding the standard deviation?
Converting the above probability distribution into a game, players can risk $4 to bet
that a seven will not be rolled in four throws of the dice. If the player wins, he will
be paid $7 (a three-dollar profit), but if the player loses he/she loses his/her $4 bet.
What is the expected value for this game? Is this a smart bet to make in the long
run?

Solution:
The standard deviation isσ = 3.50 , but this is not very useful since there are
only two outcomes for a trial. We know either a person will win 3 dollars
or will lose 4 dollars. The standard deviation is not needed to capture the
variation. It is easier to see the variation by simply viewing the win and
loss probabilities.

Thank you.

Posted to STATS 1 on Sunday, October 20, 2013   Replies: 1


Professor Mcguckian
10/21/2013
9:16 AM EST

Hi Natalie,

Please make separate posts for each question.  Otherwise, others who wish to view the response to your questions are possibly forced to read through material related to a question they do not need help with. Also, the post cannot be easily categorized for easy searching later. It also prevents me from adding a video response. For example, I would have solved your second question on video if it was posted on its own, but since I do not want to create a video that cannot be easily searched for I opted for a text response.    

For the first question, an interval using two standard deviations is standard, and that is what you should use to answer the question if it is asked of you. You can also use the z-score formula to solve the problem because it is identical to using an interval with two standard deviations above and below the mean. Remember, we interpret the z-score as being unusual if it is less than -2 or more than +2. That is equivalent to 2 standard deviations below and above the mean, so it is the same as the interval. We do not use empirical rule unless we are told that the distribution is bell shaped.    

For your second question, the problem says to use the probability distribution for problem 1 from the section. In that distribution, it says there is a probability of 0.482 that you will roll 0 sevens in four rolls. That means that the chance you roll at least one seven is 1 - 0.482 = 0.518.  If we bet $4 (for a $3 profit) that a seven is not rolled (0 sevens), we can set up a table using X and P(X).

X P(X) X*P(X) X^2*P(X)
+$3 0.482 1.446 4.338
-$4 0.518 -2.072 8.288
    -$0.626 12.626

Then we simply apply the formula: sqrRoot(12.626 - .626^2) = 3.4977... = 3.50.

Sincerely,

Professor

 

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